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2 years ago

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Craig Terrill has many _________. Press enter to interact with the item, and press tab button or down arrow until reaching the S

ubmit button once the item is selected A uniforms B opinions C skills D debts

1 answer:

iren2701 [21]2 years ago

7 0

Answer:

is there anything that is a hint? like a picture or a story?

Explanation:

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Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0

3 years ago

In a given volume of whole blood, there are more white blood cells than red blood cells.

navik [9.2K]

Answer:

its false there are fewer white blood cells

Explanation:

8 0

2 years ago

The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the

Ray Of Light [21]

Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

$E = \dfrac{kq}{r^2}$

k = 9 x 10⁹ N.m²/C²

q is the charge of the object

$q= \dfrac{Er^2}{k}$

now,

$q= \dfrac{180000\times 0.015^2}{9\times 10^9}$

q = 4.5 x 10⁻⁹ C

q = 4.5 nC

the charge on the object is equal to 4.5 nC

8 0

3 years ago

Read 2 more answers

Em um centro de distribuição uma empilhadeira carrega uma caixa de 1000kg até o caminhão responsável pelo transporte desta carga

Yakvenalex [24]

Answer:

F = 592238.09 N

Explanation:

To find the mean force you first calculate the acceleration by using the following kinematic equation:

$v^2=v_o^2+2ax\\\\v_o=0m/s\\\\v=10km/h\\\\x=8.4m\\\\$

you do "a" the subject of the equation and replace the values of the other parameters:

$a=\frac{v^2}{2x}=\frac{10000m/s}{2(8.4m)}=595.23\frac{m}{s^2}$

next, the force, by using the second Newton law is:

$F=ma=(1000kg)(595.23\frac{m}{s^2})=595238.09\ N$

hence, the force is 592238.09 N

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TRANSLATION:

Para encontrar a força média, primeiro calcule a aceleração usando a seguinte equação cinemática:

você faz "a" o assunto da equação e substitui os valores dos outros parâmetros:

Em seguida, a força, usando a segunda lei de Newton, é:

portanto, a força é 592238.09 N

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3 years ago

A large bank establishes X-bar and R charts for the time required to process applications for its charge cards. A sample of five

Mamont248 [21]

[Find the attachment]

6 0

2 years ago

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